Implicit Diﬀerentiation and the Second Derivative Calculate y �� using implicit diﬀerentiation; simplify as much as possible. x2 + 4y 2 = 1 Solution As with the direct method, we calculate the second derivative by diﬀerentiating twice. With implicit diﬀerentiation this leaves us with a formula for y �� that involves y and y � , and simplifying is a serious consideration. Recall that to take the derivative of 4y 2 with respect to x we ﬁrst take the derivative with respect to y and then multiply by y � ; this is the “derivative of the inside function” mentioned in the chain rule, while the derivative of the outside function is 8y. So, diﬀerentiating both sides of: x2 + 4y 2 = 1 gives us: 2x + 8yy � = 0. We’re now faced with a choice. We could immediately perform implicit diﬀerentiation again, or we could solve for y � and diﬀerentiate again. If we diﬀerentiate again we get: 2 + 8yy �� + 8(y � )2 = 0. In order to solve this for y �� we will need to solve the earlier equation for y � , so it seems most eﬃcient to solve for y � before taking a second derivative. 2x + 8yy � 8yy � = 0 = −2x −2x = 8y −x = 4y y� y� Diﬀerentiating both sides of this expression (using the quotient rule and implicit diﬀerentiation), we get: y �� = = y �� = (−1)4y − (−x) · 4y � (4y )2 −4y + 4xy � 16y 2 −y + xy � 4y 2 1 We now substitute −x for y � : 4y y �� = = = = y �� = −y + xy � 4y 2 −y + x −x 4y 4y 2 − y 4y · 2 4y 4y −x2 − 4y 2 16y 3 1 − 16y 3 x −x 4y (Don’t forget to use the relation x2 + 4y 2 = 1 at the end!) How can we check our work? If we recognize x2 + 4y 2 = 1 as the equation of an ellipse, we can test our equation y � = −x/4y at the points (0, 1/2) and (1, 0). At (0, 1/2), y � = −x/4y = 0 which agrees with the fact that the tangent line to the ellipse is horizontal at that point. At (1, 0) y � is undeﬁned, which agrees with the fact that the tangent line to the ellipse at (1, 0) is vertical. Once we have learned how the value of the second derivative is related to the shape of the graph, we can do a similar test of our expression for y �� . 2